Options are:
- 20
- 21
- 22
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- 24
I recently came across the above question in a competitive exam, where we get about 30 seconds to 1 minute for solving each problem. I want to if there are quick and easy methods to compute the result.
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2HINT: The number of $0$'s in a number is equal to the number of times $10$ divides it. – Marcus M Sep 15 '15 at 18:22
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@MarcusM That is not true: consider, for instance, $101$. – Théophile Sep 15 '15 at 18:33
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I think Marcus may have chosen a more accurate duplicate, in hindsight, but I'll leave mine up, just in case. – Cameron Buie Sep 15 '15 at 18:40
2 Answers
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As a general rule, the number of $0$s at the end of $k!$ is equal to $$\sum_{i=1}^\infty \left\lfloor\frac{k}{5^i}\right\rfloor$$
The above expression is a measure of how many times $5$ divides $k!$.
Arcturus
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A zero can only occur as result of multiplying 5 with 2. so consider the factors 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100 now Tolal number of 5 that can be obtained from these factor are 24.so there wiil be 24 zero.
Manoj
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Instead of writing a brand new answer, it is better practice to fix the one that's being downvoted. – Cameron Buie Sep 15 '15 at 18:36
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@michael Zero can also occur by multiplying 0 by 0(10x10=100). I didn't quiet get the solution, could you please elaborate. – Aura Wanderer Sep 15 '15 at 18:39
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@AuraWanderer,since if we have a number ending with say one zero than it must be result of presence of 5 and 2 in its factor like 20=522,so here we have a pair (5,2).similarly for number ending with two zero say 200=55222 so we have two five's here which after multiplication with 2 will make out two zero.hope this will help. – Manoj Sep 15 '15 at 18:46